1. Use a trapezoidal estimate with 3 equal subdivisions
to find
( A) 3
( B) 9
 x
3
0
2
 6 x  9  dx .
(C ) 9.5
( D) 10
( E ) 19
2. Given the following table about f  x  , estimate
 f  x
4
0
(A) 14
dx using RRAM.
x
0
1
2
3
4
f(x)
9
4
1
0
1
(D) 5
(E) 4
(B) 10
(C) 6
AP Calculus Unit 5 Day 5
Integral Definition and intro to FTC
Riemann Sums…
• Thus far, we have used rectangles and trapezoids
to APPROXIMATE area between curves and the
x-axis.
• It would be better if we could be more accurate
in our approximations.
Riemann Sum -- from Wolfram MathWorld
Notes Here:
Height of kth
rectangle
(cn,f(cn ))
(ck,f(ck ))

kth
rectangle
c1
x0=a
ck
x2
xk-1
xk
cn
xn-1
xn=b
Width of kth
rectangle=
xk
(c1,f(c1 ))
b
a
c2
x1
1. Rectangles
are used to
approximate
f ( x)dx
2. The area of
any specific
rectangle will
be:
f (ck )xk
where
f (ck )
is the height
(c2,f(c2 ))
Rectangles extending from the x-axis to intersect the curve at the
points (ck,f(ck ))
and xk is the
width.
Notes Here:
Height of kth
rectangle
(cn,f(cn ))
f (ck )xk
(ck,f(ck ))
will be positive
or negative
based on the
value of
kth
rectangle
c1
x0=a
c2
x1
ck
x2
xk-1
xk
f (ck )
cn
xn-1
xn=b
Width of kth
rectangle=
xk
(c1,f(c1 ))
3. The product
4. The sum of
these areas
can be written
as:
n
 f (c )x
k 1
(c2,f(c2 ))
Rectangles extending form from the x-axis to intersect the curve at the
points (ck,f(ck ))
k
k
Notes Here:
Height of kth
rectangle
(cn,f(cn ))
(ck,f(ck ))
kth
rectangle
c1
x0=a
c2
x1
ck
x2
xk-1
xk
cn
xn-1
xn=b
Width of kth
rectangle=
xk
(c1,f(c1 ))
5. As we
increase the
number of
rectangles the
approximation
becomes more
accurate.
(c2,f(c2 ))
Rectangles extending form from the x-axis to intersect the curve at the
points (ck,f(ck ))
This can be
written using
the notation
that is on the
next slide.
If we use an infinite number of partitions…
Add up the areas of each partition
n
EXACT  lim  f (ck )xk
n 
k 1
The number of partitions is increasing to infinity
Width of each partition
Height of partitions
Integral Notation
b
And since

f ( x)dx
represents the EXACT amount
a
We can state that
b

a
f ( x)dx  lim
n 
n
 f (c )x
k 1
k
k
Two ways to view the limit
n
lim  f (ck )xk
n 
Number of partitions goes to
infinity
k 1
n
lim  f (ck )xk
P 0
k 1
Size of partitions goes to zero
Both equal the integral
n
lim  f (ck )xk
n 
k 1
n
lim  f (ck )xk
P 0
k 1
b
  f ( x)dx
a
Formal Definition:
Let f be a function on a closed interval [a,b], let the numbers ck be
chosen arbitrarily in the subintervals [xk-1, xk]. If there exists a number
I such that
n
lim å f (ck )Dxk = I
P ®0
k =1
no matter how P and ck’s are chosen,
Then f is integrable on [a,b] and I is the definite
integral of f over [a,b].
Examples:
• Write as an integral:
n
1. lim  ck xk , partitioned between [0, 2]
2
P 0
k 1
n
lim  ck xk   x dx
2
P 0
k 1
2
0
2
Examples:
• Write as an integral:
n
1
2. lim  xk , partitioned between [1, 4]
P 0
k 1 ck
n
41
1
lim  xk   dx
1 x
P 0
k 1 ck
Examples:
• Write as an integral:
3. lim   3(mk ) 2  2mk  5  xk ,
n
n 
k 1
on the interval [1,3]
lim   3(mk )  2mk  5 xk   3x 2  2 x  5 dx
n
2
n 
k 1
3
1
Explore Properties
x
Given
g ( x)   3t dt , evaluate g(2).
2
2
2
g (2)   3t 2 dt  0
2
When upper bound =lower bound,
b
 f (t ) dt  0
a
Explain why this makes sense based on your
knowledge of what an integral represents.
x
Given
g ( x)   f (t ) dt
3
5
 f (t ) dt  7
8
 f (t ) dt  12
10
3
3
8
1. g (3) 
2. g (5) 
3. g (10) 
4. g (8)  g (5) 
 f (t ) dt  2
x
g ( x)   f (t ) dt
Given
3
5
 f (t ) dt  7
8
 f (t ) dt  12
10
3
3
8
 f (t ) dt  2
3
1. g (3)   f (t ) dt  0
3
5
2. g (5)   f (t ) dt  7
3
10
3. g (10)   f (t ) dt  12  2  14
3
8
5
3
3
4. g (8)  g (5)   f (t ) dt   f (t ) dt  12  7  5